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Kinematics In One Direction Problem

  1. Sep 19, 2004 #1
    A runner hopes to compleate the 10,000m run in less than 30.0min. After running at a constant speed for exactly 27.0min there are still 1100m to go. The runner must then accelerate at 0.20 m/s^2 for how many seconds in order to achieve the desired time.

    Ill briefly summarize what i tryed...

    Init. Velocity = (10,000-1100) /(27min * 60s) = 5.5 m/s
    Final Velocity = 1100/(180-t) since t is not know, because i dont know yet how long it will accelerate for

    i was then gonna use VF^2 = VI^2 + 2ad to find the distance it must accelerate for, and use that to find the time, this seems rather wrong, hence my confusion, but i figured id give my idea anyway
     
  2. jcsd
  3. Sep 19, 2004 #2

    Doc Al

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    Staff: Mentor

    Good. This is the speed of the runner for the first 27 minutes.
    Don't worry about the final speed, since that's not asked for. You know the distance and time for the last leg of the run, and you need to find the acceleration. The kinematic equation that relates these three variables is:
    [tex]x = v_0 t + 1/2 a t^2[/tex]
     
  4. Sep 19, 2004 #3
    I think you misunderstood the question, i need to know how long he accelerates, in order to finish in 3 mins, if you read the question more closely youll realize its not as simple as i thought, i immediatly did what you suggested, until somone else pointed out the other details
     
  5. Sep 19, 2004 #4

    Doc Al

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    Staff: Mentor

    Oops... my bad. I forgot that the acceleration was given.

    So break the remaining 3 minutes into two segments [itex]t_1[/itex] and [itex]t_2[/itex]. Assume the final speed is [itex]v_f[/itex]. So:
    [tex]v_f = v_0 + a t_1[/tex]
    [tex]t_1 + t_2 = 180[/tex]
    I'll leave the last for you: total distance traveled during [itex]t_1 + t_2[/itex] = 1100 m.

    I hope that helps.
     
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